Cryptography Reference
In-Depth Information
LEMMA 11.27
A ssu m e that the pairing
,
:
B
×
A
→
μ
n
is nondegeneratein
A
(that is, if
b, a
=1
for all
b
∈
B
,then
a
=0
).
(a) T he m ap
A
ψ
a
isinjective.
(b) If
#
A
=#
B
,then the pairing isalso nondegeneratein
B
.
→
Hom(
B, μ
n
)
given by
a
→
PROOF
Suppose
ψ
a
is the trivial homomorphism. This means that
b, a
=
ψ
a
(
b
) = 1 for all
b
∈
B
. The nondegeneracy in
A
implies that
a
=0. This
proves (a).
Let
B
1
=
{b ∈ B |b, a
=1forall
a ∈ A}.
Then each
a ∈ A
gives a well-defined homomorphism
β
a
:
B/B
1
→ μ
n
given by
β
a
(
b
mod
B
1
)=
b, a
. f
β
a
is the trivial homomorphism, then
b, a
=1forall
b ∈ B
, which means that
a
= 0. Therefore,
A
injects
into Hom(
B/B
1
,μ
n
), which has order #
B/
#
B
1
, by Lemma 11.26. Since
#
A
=#
B
,wemusthave#
B
1
=1. But
B
1
= 0 is exactly what it means for
the pairing to be nondegenerate in
B
. This proves (b).
A converse of part (b) of Lemma 11.27 holds.
LEMMA 11.28
Suppose
,
:
B
×
A
→
μ
n
is nondegenerateinboth
A
and
B
.Then
#
A
=
#
B
.Infact,
A
Hom(
B, μ
n
)
and
B
Hom(
A, μ
n
)
.
PROOF
By Lemma 11.27, we have an injection from
A
to Hom(
B, μ
n
),
so #
A
≤
#Hom(
B, μ
n
)=#
B
. Reversing the roles of
A
and
B
,wehave
#
B
#Hom(
A, μ
n
)=#
A
. Therefore, #
A
=#
B
, and the injections are
isomorphisms.
≤
LEMMA 11.29
Let
M
be a finiteabe ian group and let
α
:
M
→
M
be a hom om orphism .
Then
#Ker
α
=#
M/
#
α
(
M
)
.
PROOF
By Theorem B.6 (in Appendix B),
#
M
=(#Ker
α
)(#
α
(
M
))
.
The result follows.
The following technical lemma is the key to proving the nondegeneracy of
the Tate-Lichtenbaum pairing.
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